Theorems about the direct product of two topological spaces

A lot of results concerning the direct product (Z, \alpha, \beta) of two topological spaces X and Y are trivial if you consider the usual particular construction of these entities. I find it more satisfying to derive them directly from the universal definitions.

See also: the equivalent page on direct sums.

In the following, we will always have X and Y topological spaces and ((Z, \tau), \alpha, \beta) a direct product thereof.

No duplicates

In the Z = X \times Y construction, this is just the trivial fact that if you know (a, b), you know both a and b.

Theorem:

For every c_1, c_2 \in G, if \alpha(c_1) = \alpha(c_2) and \beta(c_1) = \beta(c_2), then c_1 = c_2.

Proof:

Let us suppose that we have c_1, c_2 \in Z such that \alpha(c_1) = \alpha(c_2) and \beta(c_1) = \beta(c_2).

Let us take Z' a singleton topological space, that is that Z' is a singleton (singletons exist!) and we equip it with the only possible topology, which is both the indiscrete and the discrete one.

We define \alpha' and \beta' as the constant mappings, Z' \to X and Z' \to Y respectively, that map the only element of Z' to \alpha(c_1), which is also \alpha(c_2), and to \beta(c_1), which is also \beta(c_2), respectively.

We define \gamma_ and \gamma_2 as the constant mappings, both Z' \to G, that map the only element of Z' to c_1 and to c_2 respectively.

Since these four mappings are constant, or, alternatively, since the topology on Z' is discrete, they are all continuous.

Furthermore, for both \gamma = \gamma_1 and \gamma = \gamma_2 we have:

    \[\alpha' = \alpha \circ \gamma\]

    \[\beta' = \alpha \circ \gamma\]

Since (Z, \alpha, \beta) is a direct product of X and Y, there can be only one morphism \gamma: Z' \to G satisfying these two relations. Hence necessarily \gamma_1 = \gamma_2, which implies c_1 = c_2. \blacksquare

Existence of an antecedent

If we have a \in X and b \in Y, does there exist a c \in Z such that \alpha(c) = a and \beta(c) = b? Again, this is trivial in the Z = X \times Y representation: c = (a, b) is the answer. Sticking to the universal definition of the direct product, we have:

For any a \in X and b \in Y, there exists c \in Z such that \alpha(c) = a and \beta(c) = b.

Proof:

We take Z' as a singleton topological space as above, and \alpha' and \beta' as the mappings Z' \to X and Z' \to Y respectively that map the unique element of Z to a and to b respectively. These two mappings again are continuous, since Z' has the discrete topology.

The universal definition of (Z, \alpha, \beta) as a direct product implies that there exists \gamma: Z' \to Z such that \alpha' = \alpha \circ \gamma and \beta' = \alpha \circ \gamma. Taking c as the image by \gamma of the unique element of Z, we thus have a = \alpha(c) and b = \beta(c). \blacksquare.

This result and the preceding one come together as:

For any a \in X and b \in Y, there exists a unique c \in Z such that \alpha(c) = a and \beta(c) = b.

The associated morphisms are surjective

At least, usually.

If X = \emptyset or Y \neq \emptyset, then \alpha is surjective. If Y = \emptyset or X \neq \emptyset, then \beta is surjective.

This is just a consequence of the general result proved in exercise 7 (corrected): in any category, if (P, \alpha, \beta) is a direct product of A and B, and if \Mor(A, B) \neq \emptyset, then \alpha is an epimorphism (and similarly, exchanging the roles of A and \alpha, and B and \beta, respectively); plus the fact that in the category of topological spaces, epimorphisms are surjective mappings.

Characterisation of the topology

That the topology on the the direct product space Z is the one generated by the collection of all \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y is a stipulation of the explicit construction of Z as X \times Y. It can however be proven without reference to this construction.

The topology of Z is the topology generated in Z by the collection of all \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y. A subset of Z is open if and only if it is the union of some collection of subsets of the form \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) with U open in X and V open in Y.

Proof:

Let \tau be the direct product topology on Z and \tau_0 the topology on Z generated in Z by the collection of all \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y.

Since \alpha and \beta are defined as continuous, necessarily \tau contains all the \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y. Hence \tau is finer than \tau_0. We wish to show that \tau_0 is also finer than \tau.

The mapping \alpha, considered from (Z, \tau_0) to X, remains continuous, since all the \alpha^\leftarrow(U) for U open in X are in \tau_0. Similarly, \beta is continuous (Z, \tau_0) \to Y. Hence we have a topological space (Z, \tau_0) and two continuous mappings \alpha and \beta, from (Z, \tau_0) to X and Y respectively. Since ((Z, \tau), \alpha, \beta) is a direct product of X and Y, there exists a unique continuous mapping \gamma: (Z, \tau_0) \to (Z, \tau) such that \alpha = \alpha \circ \gamma and \beta = \beta \circ \gamma.

This implies that any element c \in Z has the same image as \gamma(c) by both \alpha and \beta. The first theorem above implies that \gamma(c) = c. Thus \gamma = \Id_Z.

Hence \Id_Z is continous (Z, \tau_0) \to (Z, \tau); which means that \tau_0 is finer than \tau.

Hence \tau = \tau_0, that is the topology \tau of the direct product is the topology of Z generated by the collection of all \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y.

The open sets of the topology generated by a collection of sets \mathcal B are the unions of any number of finite intersections of elements of \mathcal B.

In this case, \mathcal B = \{ \alpha^\leftarrow(U) \}_{ U\ open\ in\ X } \cup \{ \beta^\leftarrow(V) \}_{ V\ open\ in\ Y }. A finite intersection of elements of \mathcal B will be of the form \bigcap_{i=1}^p \alpha^\leftarrow(U_i) \cap \bigcap_{j=1}^q \beta^\leftarrow(V_j) with the U_i‘s open subsets of X and the V_j‘s open subsets of Y. Since intersections are well-behaved respective to inverse images, and since finite intersections of open subsets are open subsets, this can be written \alpha^\leftarrow(\bigcap_{i=1}^p U_i) \cap \beta^\leftarrow(\bigcap_{j=1}^q V_j) = \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) with U open subset of X and V open subset of Y. Hence an open subset of Z is a union of subsets of Z of the form \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) with U open subset of X and V open subset of Y\blacksquare

Continuity of a mapping to a product space

If E is a topological space and \phi a mapping E \to Z, then \phi is continuous if and only if both \alpha \circ \phi and \beta \circ \phi are continuous.

If \phi: E \to Z is continuous, since both \alpha: Z \to X and \beta: Z \to Y are continuous, by composition of continuous mappings we conclude that \alpha \circ \phi and \beta \circ \phi are continuous.

Conversely, suppose that we know that \alpha \circ \phi and \beta \circ \phi are continuous. Let W be an eventual open subset of Z. We wish to show that \phi^\leftarrow(W) is an open subset of E.

Let a be an eventual point of \phi^\leftarrow(W). Then \phi(a) \in W. Since W is open in Z and (see above) “a subset of Z is openif and only if it is the union of some collection of subsets of the form \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) with U open in X and V open in Y”, there exist U open in X and V open in Y such that \phi(a) \in \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) \subseteq W. Taking \phi^\leftarrow of this expression, we obtain a \in \phi^\leftarrow(\alpha^\leftarrow(U) \cap \beta^\leftarrow(V)) \subseteq \phi^\leftarrow(W). But \phi^\leftarrow(\alpha^\leftarrow(U) \cap \beta^\leftarrow(V)) = \phi^\leftarrow(\alpha^\leftarrow(U)) \cap \phi^\leftarrow(\beta^\leftarrow(V)) = (\alpha \circ \phi)^\leftarrow(U) \cap (\beta \circ \phi)^\leftarrow(V). Since we know that both \alpha \circ \phi and \beta \circ \phi are continuous, this last expression is an open subset of E, that contains a and is included in \phi^\leftarrow(W). Thus \phi^\leftarrow(W) is a neighbourhood of a. This being the case for any a \in \phi^\leftarrow(W), \phi^\leftarrow(W) is an open subset of E. \blacksquare

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